What is f(x) = int (3x+1)^2-2x-1 dx if f(2) = 5 ?

1 Answer
Apr 15, 2018

f(x)=x^2(3x+2)-27

Explanation:

First, evaluate the integral:

color(white)=int((3x+1)^2-2x-1) dx

=int(3x+1)^2 dx -int2x dx -int1 dx

=int(3x+1)^2 dx -(2x^2)/2-x+C

=int(3x+1)^2 dx -x^2-x+C

Let u=3x+1, which means du=3dx, or dx=(du)/3:

=intu^2/3 du -x^2-x+C

=1/3intu^2 du -x^2-x+C

=1/3*u^3/3-x^2-x+C

=u^3/9-x^2-x+C

=(3x+1)^3/9-x^2-x+C

=(27x^3+27x^2+9x+1)/9-x^2-x+C

=3x^3+3x^2+x+1/9-x^2-x+C

=3x^3+2x^2+C

=x^2(3x+2)+C

Now, plug in x=2 and set this equal to 5 and solve for C:

2^2(3(2)+2)+C=5

4(8)+C=5

32+C=5

C=-27

This is our set constant. This means that our f(x) is:

f(x)=x^2(3x+2)-27

Hope this helped!